What is shown below is a reference from the text Functional Analysis by Walter Rudin.
If $X$ s a vector space over the field $\Bbb F$ then the following notation well be used $$ x+A=\{x+a:a\in A\}\\ x-A:=\{x-a:a\in A\}\\ A+B:=\{a+b:a\in A\wedge b\in B\}\\ \lambda A:=\{\lambda a:a\in A\} $$ for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.
Definition
If $\tau$ is a topology on a vector space $X$ such that
- every point of $X$ is a closed set
- the vector space operation are continuous with respect to $\tau$
we say that $X$ is a topological vector space equipped with the vector topology $\tau$.
So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.
Proposition
Let be $X$ a topological vector space over the field $\Bbb F$ and for some $v\in X$ and $\lambda\in\Bbb F$ we define the $v$-traslation and $\lambda$-multiplication functions through the condition $$ T_v(x):=x+v\,\,\,\text{and}\,\,\,M_\lambda(x):=\lambda x $$ for any $x\in X$. So the above two defined functions are homeomorphisms.
Proof. Omitted.
One consequence of the above proposition is that every vector topology $\tau$ is translation-invariant, that is a set $E\subseteq X$ is open if and only if each of its translates $v+E$ is open and thus $\tau$ is completely determined by any local base so that the open sets of $X$ are then precisely those that are unions of translates of members of any local base. In a vector space context, the term local base will always mean a local base at $0$. A local base of a topological vector space $X$ is thus a collection $\mathscr B(0)$ of tneighborhoods of $0$ such that every neighborhood of $0$ contains a member of $\mathscr B(0)$$.
Definition
Let be $X$ a topological vector space over the field $\Bbb F$ of the real or complex numbers. So a subset $C$ of $X$ is said convex if $$ tC+(1-t)C\subseteq C $$ for any $t\in[0,1]$. Then a subset $B$ of $X$ is said balanced if $$ \alpha B\subseteq B $$ for each $\alpha\in\Bbb F$ such that $|\alpha|\le 1$.
Theorem
In a topological vector space $X$ over the field $\Bbb F$ of the real or complex numbers every neighborhood of $0$ contains a balanced neighborhood of $0$ and every convex neighborhood of $0$ contains a balanced convex neighborhood of $0$.
Proof. See here.
So the last theorem can be restarted in terms of local bases so that let us say that a local base $\mathscr B$ is balanced if its members are balanced sets and let us call $\mathscr B$ convex if its members are convex sets.
Corollary
Every topological vector space has a balanced local base and every locally convex space has a balanced convex local base.
So I ask to discuss if to prove the above corollary is necessary to use the Axiom of Choice. Indeed to prove the above corollary I use the following argumentation. So the by the previous theorem it seems to me that we can claim that exist a function $f$ from the neighborhood system of $0$ in the neighborhood system of $0$ such that $V_U:=f(U)$ is a balanced neighborhood of $0$ contained in $U$ and precisely the above theorem states that we can make this function picking for any $U$ (surely we can thought that $0$ has no many finite neighborhoods!) another nneighborhood of $0$ that is balanced and here contained and to do this I do not know if we have to use the Axiom of Choice. Then clearly the last corollary is true for a local base of the null vector so I ask to explain if it is true for a local base of another different point. So could anyone help me, please?
